Let ABCD be a quadrilateral. If E and F are t | vedclass

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(C) Let the position vectors of vertices $A, B, C, D$ be $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}, \overrightarrow{d}$ respectively

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(C) Let the position vectors of vertices $A, B, C, D$ be $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}, \overrightarrow{d}$ respectively.Since $E$ is the midpoint of $AC$,$\overrightarrow{e} = \frac{\overrightarrow{a} + \overrightarrow{c}}{2}$.Since $F$ is the midpoint of $BD$,$\overrightarrow{f} = \frac{\overrightarrow{b} + \overrightarrow{d}}{2}$.The given expression is $(\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) = k \overrightarrow{FE}$.Substituting the vectors: $(\overrightarrow{b} - \overrightarrow{a} - (\overrightarrow{c} - \overrightarrow{b})) + (\overrightarrow{d} - \overrightarrow{a} - (\overrightarrow{c} - \overrightarrow{d})) = k \overrightarrow{FE}$.Simplifying: $(\overrightarrow{b} - \overrightarrow{a} - \overrightarrow{c} + \overrightarrow{b}) + (\overrightarrow{d} - \overrightarrow{a} - \overrightarrow{c} + \overrightarrow{d}) = k \overrightarrow{FE}$.$(2\overrightarrow{b} - 2\overrightarrow{a} - 2\overrightarrow{c} + 2\overrightarrow{d}) = k \overrightarrow{FE}$.$2(\overrightarrow{b} + \overrightarrow{d}) - 2(\overrightarrow{a} + \overrightarrow{c}) = k \overrightarrow{FE}$.Using $\overrightarrow{b} + \overrightarrow{d} = 2\overrightarrow{f}$ and $\overrightarrow{a} + \overrightarrow{c} = 2\overrightarrow{e}$:$2(2\overrightarrow{f}) - 2(2\overrightarrow{e}) = k \overrightarrow{FE}$.$4(\overrightarrow{f} - \overrightarrow{e}) = k \overrightarrow{FE}$.Since $\overrightarrow{FE} = \overrightarrow{e} - \overrightarrow{f}$,we have $4(-(\overrightarrow{e} - \overrightarrow{f})) = k \overrightarrow{FE}$.$-4 \overrightarrow{FE} = k \overrightarrow{FE}$.Therefore,$k = -4$.

Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the midpoints of the diagonals $AC$ and $BD$ respectively and $(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})= k \overrightarrow{FE}$,then $k$ is equal to

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